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Could someone help me, please, with this exercise?

Consider a sequence of complex numbers $\{a_n\}$ such that $a_n=a_m $ iff $ n\cong m $ mod $q$ for some positive integer $q$.

Define the Dirichlet L-series associated to the sequence by

$$L(s)=\sum_{n=1}^{\infty} \frac{a_n}{n^s} \ \ \ \text{ for Re}(s)>1. $$

Also define $$M(x)=\sum_{m=0}^{q-1}a_{q-m} e^{mx}\ \ \ \text{ with }\ \ a_0=a_q.$$

Questions

  1. How can we show that $$ L(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{M(x)x^{s-1}}{e^{qx}-1}dx, \ \ \text{for Re}(s)>1 ? $$ Note: $\Gamma(s)$ is the Gamma function.
  2. How does that imply $L(s)$ is continuable into the complex plane, with the only possible singularity a pole at $s=1$.

Any help is really appreciated.

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  • $\begingroup$ What have you tried? Do you know at least one example of this phenomenon (e.g., where every $a_n$ is 1)? $\endgroup$ – KCd Apr 1 '13 at 0:11
  • $\begingroup$ Hint: Look at the analytic continuation of the Riemann zeta function. It's pretty much that same proof, except now we're allowing for the inclusion of a Dirichlet character. $\endgroup$ – Brent J Apr 1 '13 at 2:18
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(1) Hint: write $$ \frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{M(x)x^{s-1}}{e^{qx}-1}dx = \frac{1}{\Gamma(s)} \sum_{m=0}^{q-1} \int_{0}^{\infty}\frac{a_{q-m}e^{mx}x^{s-1}}{e^{qx}} \big( 1 + e^{-qx} + e^{-2qx} + e^{-3qx} + \cdots \big) \,dx $$ and integrate term by term.

(2) Hint: in the equation you showed in part (1), write $M(x) = (M(x)-M(0))+M(0)$ and split into two integrals. The first integral should converge for any complex $s$, while the second integral will produce a pole at $s=1$.

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  • $\begingroup$ Your method in (2) will give you an analytic continuation to $\textrm{Re}(s)>0$ except a possible pole at $s=1$. $\endgroup$ – Sungjin Kim Apr 1 '13 at 15:54
  • $\begingroup$ GreGMartin Thanks for the hint. I've figured it earlier but I did not know how to do the second part. $\endgroup$ – yaa09d Apr 1 '13 at 18:28
  • $\begingroup$ In (2), how can I show that the first integral is entire? $\endgroup$ – Gobi Oct 21 '13 at 3:20

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